ratio of sidelengths wanted, equal angles given (2009 Zhautykov City VIII P4)

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ratio of sidelengths wanted, equal angles given (2009 Zhautykov City VIII P4)

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#1 h Sep 5, 2019, 6:25 AM • 1 Y

Y by Adventure10

The point $D$ is the midpoint of the side $AC$ of the triangle $ABC$ . On the side $BC$ , a point $E$ is chosen such that $\angle BEA = \angle CED$ . Find the ratio of the lengths of $AE: DE$ .

This post has been edited 1 time. Last edited by parmenides51, Dec 19, 2022, 11:59 AM

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#2 h Sep 5, 2019, 6:38 AM • 1 Y

Y by Adventure10

Answer is
2:1 most probably
Posting the solution
Let M be midpoint of AE and N be midpoint AB
As MD|| CE, MD||BC So N,M,D are collinear
Also BE||ND
M(EMD) = M(AEC)
and EC||ND
M(EDM) = M(DEC)
so by initial statement,
EMD is isosceles.
EM = MD
2×EM = 2 ×MD
AE =2 × MD
This follows that ratio is 2 :1.
........................................................................................................
I don't understand why nobody discuss a question after I post.

This post has been edited 2 times. Last edited by Shroud666, Sep 6, 2019, 6:54 AM

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#3 h Sep 11, 2019, 5:58 PM • 2 Y

Y by Adventure10, Mango247

What I did is just constructed CF parallel to DE and and extended E to meet at F so that i can use the mid point theorem. Then, it can be solved easily

This post has been edited 1 time. Last edited by dumbduck, Sep 11, 2019, 5:59 PM
Reason: Made the post more clear

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#4 h Jul 28, 2023, 7:13 PM

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parmenides51 wrote:

The point $D$ is the midpoint of the side $AC$ of the triangle $ABC$ . On the side $BC$ , a point $E$ is chosen such that $\angle BEA = \angle CED$ . Find the ratio of the lengths of $AE: DE$ .

This was too simple...
You simply locate $F\in AE$ such that $DF//BC$
Then:
$\angle BEA=\angle EFD...(I)$ $\angle DEC=\angle EDF...(II)$ By $(I)$ and $(II):$
$\Rightarrow \angle DFE=\angle FDE$ $\Rightarrow DE=FE$ Since $DF$ is the median base of $\triangle AEC:$
$\Rightarrow AF=FE=DE$ $\Rightarrow \frac{AE}{DE}=\frac{2}{1}_\blacksquare$

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